Is $\left\{x\mid x\text{ is a countable set}\right\}$ a set? I found this question in an exam:
Is $\left\{x\mid x\text{ is a  countable set}\right\}$ a set?
We are working in ${\sf ZFC}$.
 A: If the set of all countable sets existed, then by the axiom of unions, the union of all countable sets would exist. That would be the whole universe, since everything belongs to some countable set, e.g., $x\in\omega\cup\{x\}$. But there is no universal set; for every set $S$, the set $\{x\in S:x\notin x\}$ does not belong to $S$.
A: If $T_1=\lbrace x \mid x \text{ is countable} \rbrace$ is a set, so
is $T_2=\lbrace x \mid x \text{ is a singleton} \rbrace$ by the axiom
of subsets, so is $T_3=\lbrace x  \mid x \text{ is a set} \rbrace$ by the axiom of replacement,
but it is well known that $T_3$ is not a set( if it was, by
the axiom of subsets we could form $T_4=\lbrace x \mid x\not\in x\rbrace$
and this would be Russell’s paradox).
A: This axiom
can only be applied in the form
$$
\{x \in X \colon x \text{ is countable}\}
$$
for a given set $X$. So what you give is not a valid definition for a set.
However this does not itself prove that there not exists a set with the property you have stated (i.e. that contains all countable sets).
